Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 1
00 n. (b) what is the work done on the lift by the gravitational force in this process? (c) what is the total work done on the lift?
1 answer:
The weight of the elevator is
W = (1500 kg)*(9.8 m/s²) = 1.47 x 10⁴ N
Because the frictional resistance is R = 100 N, the tension in the cable for dynamic equilibrium is
T = W + R = 1.47 x 10⁴ + 100 = 1.48 x 10⁴ N
By definition, the work done in lifting the elevator by 40 m is
(1.48 x 10⁴ N)*(40 m) = 5.92 x 10⁵ J = 592 kJ
Answer: 592 kJ (or 5.92 x 10⁵ J)
W = (1500 kg)*(9.8 m/s²) = 1.47 x 10⁴ N
Because the frictional resistance is R = 100 N, the tension in the cable for dynamic equilibrium is
T = W + R = 1.47 x 10⁴ + 100 = 1.48 x 10⁴ N
By definition, the work done in lifting the elevator by 40 m is
(1.48 x 10⁴ N)*(40 m) = 5.92 x 10⁵ J = 592 kJ
Answer: 592 kJ (or 5.92 x 10⁵ J)
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OPTION A.